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let a=x^2+4. rewrite the following equation in terms of a. (x^2+4)^2+32=12x^2+48 in resulting equation what is the coefficient of the a term in the resulting equation,what is the constant

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4 votes
We're given:

a=x^2+4

And with that we must rewrite the equation
(x^2+4)^2+32=12x^2+48 in terms of
a.

Currently, the equation is written in terms of
x^2, so lets start by writing
x^2 in terms of
a:

a=x^2+4

x^2=a-4

Now lets replace every
x^2 in
(x^2+4)^2+32=12x^2+48 by
a-4:

(x^2+4)^2+32=12x^2+48

(a-4+4)^2+32=12(a-4)+48

a^2+32=12(a-4)+48

Now that we have the equation in terms of
a, lets equal it to 0:

a^2+32=12(a-4)+48

a^2+32-12(a-4)-48=0

a^2+32-12a+48-48=0

a^2-12a+32=0

So the equation (in terms of
a) is written in a standard form as
a^2-12a+32=0.

We can clearly see that the coefficient of
a is -12 and the constant is 32.
User Eandersson
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